Show form default Validation Status Programmatically

Question

I want to submit a form using ajax. So I am not using the type=submit. I am using a onClick event on a link() to send the for

Answer 1

You can do it if you let your form have a submit button and return false! And you cán do it in the same event handler as the non-submits! So, first test if you are part of a form and if so, make it check Validity and never return true (even if valid)!

$('.ajx')   
    .on("submit click", function(e) {

        var $this           = $(this)

        //Force native form validating and notification
        form = $this.closest('form')[0]
        if (form) {
            //Follow through with form submit (element must be of submit type!)
            if(!form.checkValidity()) {
                //don't ask me!
                sleep(3000);
                return false
            }
        }           

        //only preventDEfault AFTER possible form element click
        e.preventDefault()

        //...your project further ajax code
        //Makes sure your form never submits
        if (e.type=='submit') return false
}

Small downside: You have to do this on the submit button, but it's always possible to change your into type=submit. You don't have to change the non form 's though!